Problem: $\int x\cos(5x^2)\,dx\,= $ $+~C$
Solution: Notice that we can rewrite the integral as $ \int \cos(5x^2)\cdot \,x\, dx\,$. If we let $ {u=5x^2}$, then ${du=10x \, dx}$ and $x\, dx=\dfrac{du}{10}}$. Substituting gives us: $\begin{aligned}\int \cos({5x^2})\,\cdot x\,dx}\,&= \int \cos({u})\, \cdot \dfrac{du}{10}}\\\\\\ &= \int \cos(u)\cdot\dfrac{1}{10}\,du\\\\\\ &=\dfrac{1}{10} \int \cos(u)\, du\end{aligned}$ We recognize this antiderivative. $\begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx}&= \dfrac{1}{10} \int \cos(u)\, du\\\\\\ &=\dfrac{1}{10}\cdot \sin(u)+C\\\\\\\end{aligned}$ We can now substitute back to find the antiderivative in terms of $x$. ∫ e x 1 + e 2 x d x = 1 10 sin ( u ) + C = 1 10 sin ( 5 x 2 ) + C \begin{aligned}\phantom{\int\dfrac{e^x}{1+e^{2x}}dx~}&=\dfrac{1}{10}\sin(u)+C\\\\\\\ &=\dfrac{1}{10}\sin(5x^2)+C\end{aligned} The answer: $\int x\cos(5x^2)\,dx\,=\dfrac{1}{10}\sin(5x^2)+C $